77. Two equivalent perpendicularities.

by Virgil Nicula, Aug 4, 2010, 3:07 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=265876
Quote:
Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$ and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$ respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$ which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ .
Proof 1. Suppose w.l.o.g. $ b > c$ . Observe that $ \left\|\begin{array}{c} m(\angle KAB) = 90^{\circ} - C \\ \\ m(\angle KAP) = 90^{\circ} - \frac {B - C}{2} \\ \\ m\left(\angle APC\right) = 90^{\circ} - \frac {B - C}{2}\end{array}\right\|$ . Denote $ \left\|\begin{array}{c} D\in BC \\ \ AD\perp BC\end{array}\right\|$ . Therefore, $ \frac {KB}{KP} = \frac {AB}{AP}\cdot\frac {\sin\widehat {KAB}}{\sin\widehat {KAP}} =$ $ \left|\frac {c\cdot\sin (90^{\circ}- C)}{AP\cdot \cos\frac {B - C}{2}}\right| =$ $ \frac {c\cdot |\cos C|}{DC} = \frac cb$ . In conclusion, $ \frac {KB}{KP} = \frac {TB}{TC}$ $ \implies$ $ KT\parallel PC$ $ \implies$ $ KT\perp BC$ .

Proof 2 (proiectively). Denote $ L\in BC\cap AP$ , the diameter $ [XY]$ , the line $ d$ for which $ K\in d$ , $ d\parallel PC$, i.e. $ d\perp BC$ and the points $ U\in AP\cap d$ , $ V\in PT\cap LK$ , $ W\in LK\cap PC$ , $ T_1\in d\cap AX$ , $ T_2\in d\cap BC$ . Observe that $ KU=KT_1$ because $ K\in AO$ - the $ A$-median in $ \triangle AXY$ and $ \overline {UKT_1}\parallel \overline {XOY}$ . Thus, the division $ \{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the pencil $ P\{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the divison $ \{K,W;L,V\}$ is harmonically. Therefore, $ KT_2\parallel WC$ $ \Longrightarrow$ $ KT_2=KU$ $ \Longrightarrow$ $ KT_1=KT_2$ $ \Longrightarrow$ $ T_1\equiv T_2\equiv T$ $ \Longrightarrow$ $ KT\perp BC$ .
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:48 PM

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